Suppose we have an adiabatic and isothermal free expansion of an ideal gas at temperature $T_0$ so that the gas incurs an increase in entropy $$\Delta S_{sys}=Nk_B \ln\left(\frac{V_f}{V_i}\right)>0$$

The expansion occurs in adiabatic conditions so clearly $Q=0$. Also, the work done by the gas is $W=0$ because we are examining a free expansion into a vacuum. Thus we have that $\Delta U_{sys}=0$. This is all clear to me. The trouble begins when I calculate the change in helmholtz free energy $\Delta F$. The change in helmholtz free energy for this process is calculated as follows: $$\Delta F=\Delta U -T\Delta S$$ $$\Rightarrow \Delta F_{sys}=0-T_0\cdot Nk_B \ln\left(\frac{V_f}{V_i}\right) $$ $$\therefore \Delta F_{sys}<0$$

Now one of the uses of $F$ is that the negative of the change in $F$ at constant temperature tells us "the maximum work that can be extracted when a process occurs at constant temperature". We can see this from the equation $-\Delta F\geq W_{by\,the\,system}$. But if this interpretation is true, then the maximum work that can be extracted from a free expansion should be

$$W_{max\, possible}=T_0\cdot Nk_B \ln\left(\frac{V_f}{V_i}\right)$$

To be perfectly honest, this result sounds like gibberish to me. How can we possibly extract work from a free expansion into a vaccuum at constant temperature? Is my understanding of free expansions incorrect, my understanding of Helmholtz free energy incorrect, or perhaps both? Any help on this issue would be most appreciated!


1 Answer 1


It is not the maximum work that can be extracted from free expansion. It is the maximum work that can be extracted at constant temperature from the initial state to the final state by any process.

  • $\begingroup$ Thanks! Okay that makes perfect sense. Just as an aside before I accept your answer, is the temperature that appears in the definition of the H free energy ($H=U-TS$) the temperature of the surroundings, the temperature of the system, the temperature of the boundary between the two or perhaps all 3 at once? One of the texts I use (schroeder) indicates that it is the temperature of the surroundings however in Atkins' physical chemistry he introduces it in the section "concentrating on the system" which makes me think its the temp of the system. Or more likely both assuming thermal equilibrium? $\endgroup$ Feb 25, 2021 at 17:21
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    $\begingroup$ H is an equilibrium physical property (state function) of the system, so please do don't think of it in terms of any process. To evaluate the change in H, however, you need to devise a reversible path (since it contains S) between the initial and final states of the system. For a reversible process, the system temperature, boundary temperature, and surroundings temperature are all equal. $\endgroup$ Feb 25, 2021 at 22:03
  • $\begingroup$ Okay perfect. As for "For a reversible process, the system temperature, boundary temperature, and surroundings temperature are all equal", would this statement still apply for a reversible adiabatic process (say the adiabatic expansion in a carnot cycle)? The expansion occurs without any internal reversibilities and hence no entropy is generated however the temperature of the system at the midway point of the expansion is neither that of the heat source nor cold sink and so $T_{sys} \neq T_{sur}$ during the process. Should I assume the statement only applies to the start and end points? $\endgroup$ Feb 26, 2021 at 7:24
  • $\begingroup$ If I recall correctly, the Helmholtz criterion for maximum work applies specifically to a process on a closed system in contact with a constant temperature reservoir at the initial (and final) temperature of the system throughout the process. $\endgroup$ Feb 26, 2021 at 12:07

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